Algorithms | Recursion | Question 6

Output of following program?

#include<stdio.h>
void print(int n)
{
    if (n > 4000)
        return;
    printf("%d ", n);
    print(2*n);
    printf("%d ", n);
}

int main()
{
    print(1000);
    getchar();
    return 0;
}

(A) 1000 2000 4000
(B) 1000 2000 4000 4000 2000 1000
(C) 1000 2000 4000 2000 1000
(D) 1000 2000 2000 1000


Answer: (B)

Explanation: First time n=1000
Then 1000 is printed by first printf function then call print(2*1000) then again print 2000 by printf function then call print(2*2000) and it prints 4000 next time print(4000*2) is called.

Here 8000 is greater than 4000 condition becomes true and it return at function(2*4000). Here n=4000 then 4000 will again print through second printf.

Similarly print(2*2000) after that n=2000 then 2000 will print and come back at print(2*1000) here n=1000, so print 1000 through second printf.

Option (B) is correct.

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